Question:hard

What is the probability that a two-digit positive integer N has the property that the difference of N and the number obtained by reversing the order of its digits is a perfect cube?

Show Hint

The difference between a two-digit number and its digit-reversal is always \( 9(a-b) \); find which digit gaps make this a perfect cube.
Updated On: Jul 21, 2026
  • 4/45
  • 5/15
  • 6/45
  • 7/45
Show Solution

The Correct Option is D

Solution and Explanation

List every pair of digits whose difference makes $ 9 \times |a-b| $ a perfect cube, then compare the count against each printed option.

  1. 4/45: undercounts the qualifying digit pairs, so it does not match the actual count.
  2. 5/15: this reduces to 1/3, far larger than the true share of qualifying numbers.
  3. 6/45: also undercounts the seven valid pairs by one, so it is incorrect.
  4. 7/45: matches the seven pairs of digits, (0,3), (1,4), (2,5), (3,6), (4,7), (5,8) and (6,9), each giving a difference of 27, which is $ 3^3 $.

Out of the $ \binom{10}{2}=45 $ ways to choose two distinct digits for the number, exactly 7 pairs differ by 3 and give a perfect-cube difference of 27. So the probability is 7/45, matching option (d).

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